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Question: How would you cut a rectangular cake into two equal pieces when a rectangular piece has already been cut out of it? The cut piece can be of any size and orientation. You are only allowed to make one straight cut.

Answer: Simple question right? There are two possible solutions to this problem. People often overlook the easier solution to this problem. Let’s start with the easiest solution. If you make one straight horizontal cut along the height of the cake, the resulting slices are of equal sizes. But this solution may not work so well on a cake with icing. So let’s rethink.

In general, when a straight cut is made at any angle through the center of a rectangle, the resulting pieces are always of equal area. So let’s consider our situation. What if we make a straight cut such that it passes through the center of both the rectangles? Since the cut halves both the rectangles, the resulting two pieces are guaranteed to have equal area. Each piece has an area equal to half the original cake minus half the area of the missing rectangular piece. This results in two pieces of equal size, assuming the cake’s height is same at all points 🙂

Let’s make believe that the cake is of uniform rectangular shape and density (a block). Here is one simple solution… Balance the cake on a fulcrum (i.e. a piece of suspended wire or something similar) and then cut it at the fulcrum! No fancy stuff needed.

Not convinced, please post a pic of a rectangle with another rectangle anywhere inside it that you think cannot be bisected by method 2.
I will then show you how to bisect it. No matter where the centers of the two rectangles lie, you can always find a line that fits both points. Even if those centers lie in empty “non cake” space. 🙂

Cake is 3 dimensional rectangular shape having length(x), width(y) and height(h)
If Cake is cut through the middle of h(height), then we have two half of cakes, it does not matter from where rectangular piece was cut

I am unhappy with this solution. As “Very Unconvinced” said, what if the slice is from within the cake?
I like his idea of using mass.

The following assumes that the cut region is inclusive to the cake, and does not initially leave the cake in two pieces.
Let the piece be of dimensions a,b with any rotation and have a centroid at vector P.
Let the original cake be of dimensions m,n and have a centroid at vector C.
After the cut, the cake’s new centroid is vector N = (mnC-abP)/(mn-ab).

Any cut through this new centroid N will have equal mass on either side of the cut.
But to make sure the remaining cake is cut into two pieces you need to follow a simple rule.

If any of the cut is to go through the void of the original piece,
it must also go though any shared border between the piece and the cake.
A shared border can be either a side or a point (if the piece was rotated).
You don’t have to worry about the case of two points of intersection since
we assumed the cake would still be in one piece before we make our cut.

If no such border exists, cut anyway you like through N. It wont matter as the piece is convex.

It’s not hard to have the initial missing rectangle intersect that straight cut, thus making it actually two straight cuts that result in 3 pieces. (For example, have the initial missing rectangle be most of the cake, leaving a thin L for the two remaining rectangles.)

There is no diagram provided with this picture, and so we have to assume that “any size and orientation” means the cut out rectangle could be anywhere. The current solution seems to indicate that the remaining could can be mentally divided into two disjoint rectangles, but this is only possible if the missing piece contains one of the original corners of the cake.

Consider what happens if a piece is removed from the center of the cake. Then the minimum number of disjoint rectangles into which the cake can be divided is four! In this case, it is impossible for a straight cut to contain the centers of all the rectangles, so the given solution cannot work.

We can make this even more tricky! Let’s assume the piece cut out of the center is a rectangle rotated by 36 degrees with respect to the orientation of the outer boundary of the cake. Now there is no way to mentally divide it into disjoint rectangles.

As you can see, a solution to this problem must be extremely general, and the simplest one is this: Pick up the cake and balance the tray on one finger. Mark the balance point by poking your finger in the icing directly above it. Set the cake down and make any straight vertical cut through the middle of this indentation.

This technique assures that both parties will get the same amount of cake (by mass or by volume) under the assumption that the cake has uniform density and is the same height everywhere.

But we can be more general and remove these assumptions! In fact, there is a theorem that asserts that there exists a plane cut through the center of mass which evenly divides the mass no matter how it is distributed!

Lastly, I want to mention in passing that if one can compute the mass distribution of the cake (any cake, period) any project it directly onto any line, then one can find a straight cut perpendicular to that line that evenly divides the cake by solving a single equation in one variable. (Said cut may not go through the center of mass however.) This would be the preferable method of dividing our uniform rectangular cake with removed parallaxial rectangle, since the mass distribution can be determined just by looking at it. No need to even pick the cake up!